## Homework 14.5 part d

Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### Homework 14.5 part d

For letter d) in problem 14.5 (6th edition), the answe key says to multiply the first half-reaction by 3, but then the new reaction contains 4P4, not 3P4.

Is this a mistake in the answer key or the actual answer? If it’s correct, could someone explain why?

Lauren Ho 2E
Posts: 70
Joined: Tue Oct 09, 2018 12:16 am

### Re: Homework 14.5 part d

The Solution Manual Errors PDF on Professor Lavelle's website doesn't point to any errors in the solution manual for problem 14.5, so I would assume that the given answer is right

Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### Re: Homework 14.5 part d

I checked the Errors PDF as well, but I still have no idea how they got 4P4 by multiplying by 3.

204929947
Posts: 76
Joined: Fri Apr 06, 2018 11:03 am

### Re: Homework 14.5 part d

Also, in part b, how come when they balance the equation out, they get rid of Br2 and only write 6OH- --> 5Br-(aq)+BrO3-(aq)+3H2O(l) ???
how did they cancel out Br2??

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

### Re: Homework 14.5 part d

MadisonFuentes1G wrote:For letter d) in problem 14.5 (6th edition), the answe key says to multiply the first half-reaction by 3, but then the new reaction contains 4P4, not 3P4.

Is this a mistake in the answer key or the actual answer? If it’s correct, could someone explain why?

The overall redox reaction has 4P4 because you multiply the oxidation half-reaction by 3, giving 3P4, and add it with the reduction half-reaction that has 1P4. Therefore in total there is 4P4. I have the 7th edition so I'm assuming the solution manual answer hasn't changed from 6th to 7th. Hope this helps!